Introductory Statistics
Lesson 4.1 B
Objective:
SSBAT construct a discrete probability distribution and
its graph.
SSBAT determine if a distribution is a probability
distribution.
Standards: S2.5B
Review:
Discrete Random Variable
 A random variable that has a finite or countable
number of possible outcomes
 The outcomes can be listed
 The outcomes can be shown as just points on a
number line
 Examples:
The number of students in a class
Discrete Probability Distribution
 Lists each possible value the variable can be and its
corresponding Probability
 Must satisfy each of the following conditions
1. Each probability is between 0 and 1, inclusive
2. The sum of all the probabilities is 1
Example of a Discrete Probability Distribution
Days with Rain, x Probability P(x)
0
0.216
1
0.432
2
0.288
3
0.064
Determine if each is a Discrete Probability Distribution.
Explain why or why not.
 Each individual probability, P(x), has to be between 0 and 1, inclusive
 The sum of the probabilities has to equal 1
1.
x
5
6
7
8
P(x)
0.28
0.21
0.43
0.15
 No – The sum of all the probabilities is 1.07 not 1
Determine if each is a Discrete Probability Distribution.
Explain why or why not.
 Each individual probability, P(x), has to be between 0 and 1, inclusive
 The sum of the probabilities has to equal 1
2.
x
1
2
3
P(x)
0.15
0.36
0.49
 Yes – The sum of all the probabilities is 1 AND
each individual probability is between 0 & 1
Determine if each is a Discrete Probability Distribution.
Explain why or why not.
 Each individual probability, P(x), has to be between 0 and 1, inclusive
 The sum of the probabilities has to equal 1
3.
x
1
2
3
4
P(x)
0.5
0.25
0.75
-0.5
 No – The probability of 4 is Negative
Find the missing value in the probability distribution:
x
1
2
P(x) 0.16 0.22
3
0.28
4
0.2
5
 Remember that all the probabilities should add
up to be 1
0.16 + 0.22 + 0.28 + 0.2 = 0.86
1 – 0.86 = 0.14
The missing value is 0.14
Constructing a DISCRETE Probability Distribution
1. Make a Frequency Distribution for the possible
outcomes
2. Find the Sum of the frequencies
3. Find the probability of each possible outcome
 Divide the Frequency of each by the sum of
the frequencies
4. Check that each probability is between 0 and 1,
inclusive, and that the sum is 1.
Example 1
An industrial psychologist administered a personality inventory test
for passive-aggressive traits to 150 employees. Individuals were
given a score from 1 to 5, where 1 was extremely passive and 5
was extremely aggressive. A score of 3 indicated neither trait.
The frequency is shown below.
Score, x
Frequency
1
24
2
33
3
42
4
30
5
21
Example 1 Continued
1st: Find the probability of each score  Divide each individual
score’s frequency by the total of the frequency column
Total of Frequencies = 150
P(1) = 24/150 = 0.16
P(2) = 33/150 = 0.22
P(3) = 42/150 = 0.28
P(4) = 30/150 = 0.20
P(5) = 21/150 = 0/14
Score, x
Frequency
1
24
2
33
3
42
4
30
5
21
Example 1 Continued
Create the Probability Distribution using a table:
x
1
2
3
4
5
P(x)
0.16
0.22
0.28
0.20
0.14
Example 2
A company tracks the number of sales new employees make each
day during a 100-day probationary period. The results for one new
employee are shown in the table below.
Construct a Discrete Probability Distribution.
Sales per day
x
Number of Days,
f
0
16
1
19
2
15
3
21
4
9
5
10
6
8
7
2
Example 2 Continued
1st: Find the Probability of each x value
Total of Frequencies = 100
Sales per day
x
Number of Days,
f
P(0) = 16/100 = 0.16
0
16
1
19
P(1) = 19/100 = 0.19
2
15
3
21
4
9
5
10
6
8
7
2
P(2) = 15/100 = 0.15
P(3) = 21/100 = 0.21
P(4) = 9 /100 = 0.09
P(5) = 10/100 = 0.10
P(6) = 8/100 = 0.08
P(7) = 2/100 = 0.02
Example 2 Continued
Create the Probability Distribution using a table:
x
0
1
2
3
4
5
6
7
P(x)
0.16
0.19
0.15
0.21
0.09
0.10
0.08
0.02
Complete Worksheet 4.1B